The Art and Craft of Problem Solving

284 CHAPTER 8 GEOMETRY FOR AMERICANS

MN; let's call thisX. Then we will need to show that

QX ·XP=MX ·XN.

5. Use the converse of Ptolemy's Theorem (8.4.30 below) to show that the four

points are concyclic. To do so, we must show that

QM ·NP+MP·NQ = QP·MN.

All of these ideas have merit, but the second and third demand many more lines to

be drawn in, and that's too confusing for now. Likewise, Ptolemy's Theorem seems

pretty complicated.

On the other hand, option #1 looks very promising, as it does seem natural that one

of the points already labeled may in fact be the center. If M, N, P, Q are concyclic, then

MN and PQ will be chords in some circle, and by the relationship between chords and

radii (8.2. 13 ), the perpendicular bisector of these chords will pass through the center

of the circle.

Notice that AB is the perpendicular bisector of MN, since MN is perpendicular to

AB (because M and N are points on the altitude perpendicular to AB), and MN is a

chord of circle i»2 (whose diameter is AB). Likewise, AC is the perpendicular bisector

ofQP.

Hence, the only viable candidate for center is A. We know that AQ = AP and

AM = AN. We have reduced the problem to showing that AM = AP. Notice that AM

and AP are each legs of right triangles inscribed in COl and i»2, respectively. Here is a

diagram showing this, eliminating some of the prior lines to avoid clutter.

A

B

The problem reduces to something quite doable. Using similar triangles, or re­

membering 8.3. 10 , we have

AM^2 =AC'·AB

and

AP^2 =AB'·AC.

Since C' and B' are the feet of altitudes of triangle ABC, they have fixed locations

and their lengths can be calculated with routine trigonometry: AC' = AC cosA and